Find the center and foci of ? 9x²+4y²-36x+24y+36=0

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Feb 9, 2018

Answer:

Center is#(2,-3)# and focii are #(2,-3+sqrt5)# and #(2,-3-sqrt5)#.

Explanation:

Let us write #9x^2+4y^2-36x+24y+36=0# in general form. In general form equation is #(x-h)^2/a^2+(y-k)^2/b^2=1#, where #(h,k)# is center of ellipse and larger of #a# and #b# is half the major axis and smaller one is half the minor axis.

The equation can be written as

#9(x^2-4x)+4(y^2+6y)=-36#

or #9(x^2-4x+4)+4(y^2+6y+9)=-36+36+36#

or #9(x-2))^2+4(y+3)^2=36#

or #(x-2)^2/2^2+(y+3)^2/3^2=1#

Hence center of ellipse is #(2,-3)#, major axis is #6# parallel to #y#-axis and minor axis is #4# parallel to #x#-axis.

Hence, we get eccentricity #e=sqrt(1-4^2/6^2)=sqrt(1-4/9)=sqrt5/3#

and focii are at a distance of #be# i.e. #3xxsqrt5/3# i.e. #sqrt5# on either side of center parallel to major axis i.e.

focii are #(2,-3+sqrt5)# and #(2,-3-sqrt5)#.

graph{(9x^2+4y^2-36x+24y+36)((x-2)^2+(y+3)^2-0.03)((x-2)^2+(y+3+sqrt5)^2-0.03)((x-2)^2+(y+3-sqrt5)^2-0.03)=0 [-6, 10, -7, 1]}

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